User tony pantev - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T10:25:59Z http://mathoverflow.net/feeds/user/439 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120697/hitchin-fibration-outside-of-type-a/120834#120834 Answer by Tony Pantev for Hitchin fibration outside of type A Tony Pantev 2013-02-05T05:32:56Z 2013-02-05T13:53:41Z <p>Here is a very brief explanation of what is going on. You can find more details in the papers that Ana lists and in arxiv.org/abs/math/0604617 as you say. </p> <p>If $G$ is a reductive group, the Hitchin base $\text{Hitch}_{G}$ is defined to be the moduli of cameral covers of $X$. Concretely</p> <p><code>$$\tag{1} \text{Hitch}_{G} = H^{0}(X, tot(\mathfrak{t}\otimes \omega_{X})/W)$$</code></p> <p>where $\mathfrak{t}$ is a Cartan subalgera in $\mathfrak{g}$. This space is non-canonically isomorphic to </p> <p><code>$$\oplus_{i = 1}^{r} H^{0}(X, \omega_{X}^{\otimes d_{i}})$$</code></p> <p>where $d_{1}, \ldots, d_{r}$ are the degrees of the members of a basis of homogeneous invariant polynomials on $\mathfrak{g}$. </p> <p>It is better to work with the description (1) since it does not involve choices. Given a section $b \in H^{0}(X, tot(\mathfrak{t}\otimes \omega_{X})/W)$ we can construct a cameral cover $\widetilde{X}_{b} \to X$ as the pull back of the natural $W$-cover </p> <p><code>$$tot(\mathfrak{t}\otimes \omega_{X}) \to tot(\mathfrak{t}\otimes \omega_{X})/W$$</code> </p> <p>by the map $b : X \to tot(\mathfrak{t}\otimes \omega_{X})/W$. </p> <p>The <em>universal cameral cover</em> then is a subvariety</p> <p><code>$$\widetilde{X} \subset tot(\mathfrak{t}\otimes \omega_{X}) \times \text{Hitch}_{G}$$</code></p> <p>(rather than a subvariety in <code>$tot(\omega_{X})\times \text{Hitch}_{G}$</code>).</p> <p>Furthermore the fibers of the Hitchin map are not the relative Picard varieties along the fibers of the universal cameral cover. They are generalized Prym varieties associated with this relative Picard varieties and the $W$-action. Roughly you will have </p> <p><code>$$T^{*}\text{Bun}^{0}_{G} = \text{Hom}(\Lambda,Pic(\widetilde{X}^{0}/\text{Hitch}^{0}_{G}))^{W}$$</code></p> <p>where $\Lambda$ is the character lattice of a maximal torus of $G$. </p> <p>However this description is only correct up to isogeny. The precise description is given in the Donagi-Gaitsgory paper and is also recalled in our paper arxiv.org/abs/math/0604617.</p> http://mathoverflow.net/questions/102802/quasi-unipotent-monodromy-for-general-families/102847#102847 Answer by Tony Pantev for Quasi-unipotent monodromy for general families Tony Pantev 2012-07-22T00:47:23Z 2012-07-23T02:36:10Z <p>Quasi-unipotency is a well defined notion at any point of the discriminant. If we have a proper family $f : X \to S$ of varieties with a smooth total space and a smooth base, and if $p \in D \subset S$ is a point of the discriminant, then we say that the local monodromy of the family near $p$ is quasi-unipotent if we can find a small analytic neighborhood $p \in U \subset S$ of $p$ in $S$, so that if $o \in U - D$ is a base point, then the monodromy representation $mon : \pi_{1}(U-D,o) \to GL(H^{i}(X_{o},\mathbb{C})$ has an image whose Zariski closure $G$ is a quasi-unipotent linear algebraic group (that is, the quotient of $G$ by its unipotent radical is a finite-group). </p> <p>In general it is rare for the local monodromy to be quasi-unipotent. If $p$ happens to be a very singular point of the discriminant, then the local monodromy tends to be big and is often as big as it can be, and not quasi-unipotent at all. However, if $p$ is at worst a normal crossing singularity of $D$, then the local monodromy is quasi-unipotent.</p> http://mathoverflow.net/questions/100943/examples-of-eigensheaves-outside-of-langlands/100953#100953 Answer by Tony Pantev for Examples of Eigensheaves outside of langlands Tony Pantev 2012-06-29T16:16:28Z 2012-06-29T16:16:28Z <p>I am not sure if you will count this but you have the examples from the other side of geometric Langlands. On any smooth variety the skyscraper sheaves of points are eigensheaves for the tensorization endofunctors of the derived category of quasi-coherent sheaves. By definition the tensorozation endofunctors are the functors corresponding to tensoring with a fixed vector bundle. These endofunctors are given by kernels (the pushforward of the vector bundle by the diagonal map) and the structure sheaves of points are an orthogonal basis of the derived category. </p> http://mathoverflow.net/questions/99287/genus-two-pencil-in-k3-surface/99291#99291 Answer by Tony Pantev for Genus two pencil in K3 surface Tony Pantev 2012-06-11T12:44:05Z 2012-06-11T12:44:05Z <p>You will get such a pencil of genus two curves provided that the base point of the pencil of lines is not one of the nine points that you blew up to get your $E(1)$. </p> <p>But if the goal was to construct a pencil of genus two curves with two base points, this model for a K3 seems to be too complicated. Why not take a K3 which is the double cover of the plane branched at a smooth sextic? There again if you pull back a pencil of lines in the plane, you will get a pencil of genus two curves with two base points. </p> http://mathoverflow.net/questions/99200/question-on-k3-surface/99216#99216 Answer by Tony Pantev for Question on K3 Surface Tony Pantev 2012-06-10T04:55:30Z 2012-06-10T04:55:30Z <p>Do you want your K3 to be smooth? In that case the answer is <em>no</em>. For the double cover to have a trivial canonical class you will have to chose a branch divisor which is a section in half of the anti-canonical class. But the anti-canonical class of the rational elliptic surface is the class of the fiber and so is never divisible by two. </p> <p>The closest you can get to such a double cover is when use an Enriques surface as the base surface. Choose a genus one pencil on your Enriques surface. It gives you a genus one fibration whose Jacobian fibration is a rational elliptic surface. However on the Enriques, the fibration has two double fibers. If you take the root double cover that is branched at the two double fibers, it has two curves of nodal singularities. When you normalize this double cover you get a K3 with an elliptic fibration which is a pullback of the genus one fibration on the Enriques. </p> http://mathoverflow.net/questions/98403/computing-chern-classes-for-products-of-varieties/98481#98481 Answer by Tony Pantev for Computing chern classes for products of varieties Tony Pantev 2012-05-31T12:20:35Z 2012-05-31T12:20:35Z <p>It appears that you are assuming that your varieties $C_{i}$ are smooth (you seem to assume that since you are talking about the tangent <em>bundle</em>). In this case each $C_{i}$ is an elliptic curve (I guess, this is what you meant by "toric") and so $C_{1}\times C_{2}\times C_{3}$ is a three dimensional abelian variety. Since it is a group its tangent bundle is trivial and all of its Chern classes are zero. For any other bundle, it will depend on how the bundle is defined. If you have more information about your other bundle it should be easy to figure out the answer.</p> http://mathoverflow.net/questions/95353/is-this-sequences-of-complexes-of-sheaves-exact/95365#95365 Answer by Tony Pantev for Is this Sequences of Complexes of Sheaves Exact? Tony Pantev 2012-04-27T14:54:14Z 2012-04-27T18:56:00Z <p>This is not related to sheafification. The sheaf $\mathbb{C}^{*}$ of locally constant functions on $M$ is already a sheaf, so sheafification will not change it. </p> <p>This sequence is not an exact sequence of complexes but it is an <em>exact triangle</em> of complexes. That is - it is an exact sequence of complexes, up to quasi-isomorphism. The obvious short exact sequence of complexes is <code>$0 \to \mathbb{C}^{*} \to \left[\begin{array}{c} \underline{\mathbb{C}}^{*} \\ \downarrow \\ A^{1}_{M} \end{array} \right] \to \left[ \begin{array}{c} A^{1}_{M,cl} \\ \downarrow \\ A^{1}_{M} \end{array}\right] \to 0 .$</code> Now note that the last complex has an obvious surjective map <code>$\left[ \begin{array}{c} A^{1}_{M,cl} \\ \downarrow \\ A^{1}_{M} \end{array}\right] \to \left[ \begin{array}{c} 0 \\ \downarrow \\ A^{2}_{M,cl} \end{array} \right]$</code> and that this surjective map is a quasi-isomorphism. So up to a quasi-isomorphism, you can replace the last term in the short exact sequence with the complex you wanted.</p> http://mathoverflow.net/questions/89683/how-to-compute-mathcalexti-x-mathcalo-y-1-mathcalo-y-2/89869#89869 Answer by Tony Pantev for How to compute $\mathcal{Ext}^{i}_{X}(\mathcal{O}_{Y_{1}},\mathcal{O}_{Y_{2}})$? Tony Pantev 2012-02-29T13:47:45Z 2012-02-29T13:47:45Z <p>If $c$ is the codimension of $Z$ in $Y_{2}$, then <code>$$\mathcal{E}xt^{i}_{X}(\mathcal{O}_{Y_{1}},\mathcal{O}_{Y_{2}}) = \wedge^{c} N_{Z/Y_{2}}\otimes \wedge^{i -c} \left( N_{Z/X}/N_{Z/Y_{2}}\right).$$</code> You can see this by Hom-ing one Koszul complex into another. You can find the calculation e.g. in Appendix A.3 of <a href="http://arxiv.org/pdf/hep-th/0208104.pdf" rel="nofollow">this paper</a> of Katz and Sharpe. </p> http://mathoverflow.net/questions/72998/finite-fundamental-groups-of-3-dimensional-calabi-yau-manifolds/73005#73005 Answer by Tony Pantev for Finite fundamental groups of 3-dimensional Calabi-Yau manifolds Tony Pantev 2011-08-16T17:32:21Z 2011-08-17T00:00:27Z <p>This intuition seems to be only loosely right. There are many smooth compact CY threefolds with large fundamental groups. For instance $\mathbb{Z}/3\times \mathbb{Z}/3$, $\mathbb{Z}/8\times \mathbb{Z}/8$, are allowed fundamental groups and I am pretty sure that those do not act freely on $S^{3}$. </p> <p>More to the point - the Calabi-Yau threefolds that have these fundamental groups are explicitly constructed and we have a pretty good idea of the shape of (at least one of) their slag torus fibrations. For instance, in the first case the Calabi-Yau fibers by genus one curves over a rational elliptic surface, and the slag fibration is compatible with the genus one fibration. In the second case the Calabi-Yau fibers by abelian surfaces and again the slag fibration is compatible. So guided by the holomorphic picture you can easily imagine a situation where you group acts freely on the CY, preserves the slag torus fibration, and the induced action on the base of the fibration is <em>not</em> free. The only thing you can conclude really is that the action of the group on any fiber sitting over a fixed point in the base is free. This is possible to arrange on a torus by taking action by translations.</p> <p>So, even if your fundamental group happens to admit some free action on $S^{3}$, this doesn't mean that the action on the base of the slag fibration will be free. And, in general, I don't expect it to be free.</p> http://mathoverflow.net/questions/68356/orthogonal-complements-of-root-lattices-in-e-8/68368#68368 Answer by Tony Pantev for Orthogonal Complements of Root Lattices in E_8 Tony Pantev 2011-06-21T12:11:22Z 2011-06-21T12:11:22Z <p>You can get different orthogonal complements for different embeddings. There are two different embeddings of $A_{7}$ in $E_{8}$ so that for the first embedding the orthogonal complement is the lattice $A_{1}$, and for the second embedding the orthogonal complement is the lattice $\langle 8 \rangle$. </p> http://mathoverflow.net/questions/62492/non-kahler-manifolds-and-the-ddc-lemma/62573#62573 Answer by Tony Pantev for Non-Kahler manifolds and the dd^c-lemma Tony Pantev 2011-04-21T21:19:15Z 2011-04-22T11:55:32Z <p>Here is an example of a Moishezon manifold which is easy to visualize. Take a high degree (e.g. a quintic) hypersurface $Z$ in $\mathbb{P}^{4}$ which has a single ordinary double point. Let $X$ be a small resolution of $Z$. Explicitly, a small analytic neighborhood of the singularity can be identified with the vertex of a cone over a two dimensional quadric and you just need to blow-up the Weil divisor which is the preimage of one ruling. The threefold $X$ is compact complex manifold and does not admit any Kaehler structure. The last statement follows for instance from a theorem of Smith-Thomas-Yau which states that a threefold with a single node will admit a symplectic small resolution only if the three sphere that vanishes at the node is homologous to zero. The high degree condition on $Z$ ensures that the vanishing cycle is not homologous to zero, hence the statement. </p> http://mathoverflow.net/questions/51965/is-there-a-quaternionic-algebraic-geometry/52027#52027 Answer by Tony Pantev for Is there a quaternionic algebraic geometry ? Tony Pantev 2011-01-14T02:44:14Z 2011-01-14T02:44:14Z <p>Take a look at Dominic Joyce's paper <a href="http://arxiv.org/abs/math/0010079" rel="nofollow">"A theory of quaternionic algebra, with applications to hypercomplex geometry"</a>.</p> http://mathoverflow.net/questions/51422/pushing-complex-structure-forward/51438#51438 Answer by Tony Pantev for Pushing Complex Structure Forward Tony Pantev 2011-01-07T21:43:49Z 2011-01-08T20:07:28Z <p>This seems to be a question about holomorphicity of diffeomorphisms in a given complex structure. Replace your covering map $E \to B$ by its Galois closure (= frame bundle) $X \to B$. Now by construction $X \to B$ is a covering space which is Galois with Galois group $G$ (= groups of self bijections of a fixed fiber of $E \to B$). Since $X \to B$ factors through $E$, every complex structure on $E$ will induce a complex structure on $X$, and a complex structure on $B$ makes $E \to B$ holomorphic if and only if it makes $X \to B$ holomorphic. But the later question is just the question of whether all elements of $G$ which act as diffeomorphisms of $X$ will preserve the complex structure. Some of them preserve it automatically, e.g. the elements of the subgroup $H \subset G$ for which $E = X/H$. But for the rest it is an actual condition. If all those diffeomorphisms preserve your complex structure, then the quotient exists as a complex manifold. If one of them doesn't, then your are out of luck. </p> <p>I don't think you can get more concrete obstructions. </p> http://mathoverflow.net/questions/47735/reference-for-the-hodge-bundle/47740#47740 Answer by Tony Pantev for Reference for the Hodge Bundle Tony Pantev 2010-11-30T01:01:04Z 2010-11-30T01:16:52Z <p>The list of references is way too long. Here are some classical texts containing both the setup and calculations:</p> <p>1) P. Deligne, Le d´eterminant de la cohomologie, Current Trends in Arithmetical Algebraic Geometry,Contemp. Math., no. 67, AMS, Providence, 1987.</p> <p>2) Gerd Faltings, Ching-Li Chai, Degenerations of abelian varieties, Springer-Verlag, 1990.</p> <p>3) L. Moret-Bailly, Pinceaux de vari´et´es ab´eliennes, Ast´erisque 129 (1985).</p> <p>And here are a couple of more recent papers that deal with the self-intersection and cohomology calculations:</p> <p>4) <a href="http://arxiv.org/pdf/alg-geom/9604017v2" rel="nofollow">http://arxiv.org/pdf/alg-geom/9604017v2</a>.</p> <p>5) <a href="http://arxiv.org/pdf/alg-geom/9703021v2" rel="nofollow">http://arxiv.org/pdf/alg-geom/9703021v2</a>.</p> http://mathoverflow.net/questions/47660/symmetric-sequence-of-blow-ups-for-the-fulton-macpherson-compactification/47696#47696 Answer by Tony Pantev for Symmetric sequence of blow-ups for the Fulton-MacPherson compactification Tony Pantev 2010-11-29T16:42:04Z 2010-11-29T20:45:12Z <p>There is a slightly bigger compactification of the configuration space constructed by Ulyanov in <a href="http://arxiv.org/pdf/math/9904049v2" rel="nofollow">http://arxiv.org/pdf/math/9904049v2</a>. It dominates the Fulton-MacPherson compactification and it is again constructed inductively. It has the advantage of the blow-ups being symmetric on each stage. There is also a symmetric construction of the Fulton-MacPherson compactification that was pointed out by Dylan Thurston. You can find a brief description of a real version of this construction in section 3 of <a href="http://arxiv.org/pdf/math/9901110v2" rel="nofollow">http://arxiv.org/pdf/math/9901110v2</a>.</p> http://mathoverflow.net/questions/46752/is-the-cotangent-bundle-to-a-kahler-manifold-hyperkahler/46755#46755 Answer by Tony Pantev for Is the cotangent bundle to a Kahler manifold hyperkahler? Tony Pantev 2010-11-20T17:42:29Z 2010-11-20T17:42:29Z <p>Such hyper Kaehler metrics do exist near the zero section, e.g. in a formal or an analytic tubular neighborhood of the zero section. After that one can use some homogeneity to spread them on the whole cotangent bundle but typically the resulting metrics are non-complete. One gets nice global metrics on the cotangent bundles of Hermitian symmetric spaces but this is pretty much it. This question was studied extensively. There are two different proofs of the existence: in <a href="http://www.google.com/url?sa=t&amp;source=web&amp;cd=4&amp;sqi=2&amp;ved=0CCMQFjAD&amp;url=http%3A%2F%2Fpeople.maths.ox.ac.uk%2Fhitchin%2Fhitchinstudents%2Ffeix.ps.gz&amp;rct=j&amp;q=Birte%20Feix&amp;ei=ogXoTKeTFsnMswahpLyyCw&amp;usg=AFQjCNEfkq4jK0h2fTxjUgWcH1_KS-uC8A&amp;cad=rja" rel="nofollow"> this work </a> of Birte Feix and <a href="http://arxiv.org/abs/math/0011256" rel="nofollow"> this work </a> of Dima Kaledin.</p> http://mathoverflow.net/questions/45116/total-space-of-the-line-bundle-mathcalo1-over-mathbbpn/45121#45121 Answer by Tony Pantev for Total space of the line bundle $\mathcal{O}(1)$ over $\mathbb{P}^n$ Tony Pantev 2010-11-06T23:59:55Z 2010-11-06T23:59:55Z <p>It is the complement $\mathbb{P}^{n+1} - {x}$ of a point in a projective space. </p> http://mathoverflow.net/questions/44634/special-fiber-of-the-neron-model-of-an-abelian-scheme-in-terms-of-limit-hodge-str/44675#44675 Answer by Tony Pantev for Special fiber of the Neron Model of an Abelian scheme in terms of Limit Hodge Structure Tony Pantev 2010-11-03T13:54:31Z 2010-11-03T13:54:31Z <p>Take a look at this <a href="http://archive.numdam.org/article/PMIHES_1983__58__5_0.pdf" rel="nofollow">classical paper of Clemens</a>. Maybe this is what you are looking for?</p> http://mathoverflow.net/questions/2548/albanese-schemes-when-does-an-initial-abelian-scheme-exist-under-a-given-sch/2606#2606 Answer by Tony Pantev for "Albanese" schemes: When does an "initial abelian scheme" exist under a given scheme? Tony Pantev 2009-10-26T12:54:18Z 2010-08-11T16:38:02Z <p>The construction of an Albanese scheme and an Albanese map for proper and geometrically irreducible schemes over a perfect field goes back to the work of Chevalley, to <a href="http://archive.numdam.org/article/SCC_1958-1959__4__A10_0.pdf" rel="nofollow">this talk</a> of Serre, and to Grothendieck, e.g. <a href="http://www.math.jussieu.fr/~leila/grothendieckcircle/FGA.pdf" rel="nofollow"> Theorem 3.3. in FGA Exp.VI </a>. The idea is always to look at the dual of an appropriate component of the Picard scheme. One just has to pile enough conditions on the setup to ensure that the Picard functor is representable. </p> <p>One thing that should be said here is that the Albanese scheme is not in general an abelian scheme but only a torsor over a semi-abelian scheme. Also if we drop the properness condition or consider our original scheme to be defined over a base scheme things become more interesting. In full generality it is possible to define an Albanese 1-motive over the base scheme (it is a complex of sheaves of abelian groups of small amplitude with typically representable cohomology) which has the desired universal property. There are various techniques for construction of this derived version of the Albanese scheme. Some use characteristic zero, resolution of singularities, and Nagata compactifications, and some use simplicial scheme resolutions. There are many cool works in this direction, e.g. the paper of <a href="http://arxiv.org/pdf/math/9906165v1" rel="nofollow"> Barbieri-Viale and Srinivas </a>, and the more recent papers of Niranjan Ramachandran and <a href="http://arxiv.org/pdf/math/0607738v2" rel="nofollow"> Ayoub and Barbieri-Viale </a>. The appendix of Mochizuki's paper mentioned in Lars' post is also excellent.</p> http://mathoverflow.net/questions/31982/simple-kahler-manifolds/31992#31992 Answer by Tony Pantev for "Simple" Kahler manifolds Tony Pantev 2010-07-15T12:33:34Z 2010-07-15T12:33:34Z <p>This is a very interesting class of manifolds which, to my knowledge, has not been studied in any detail. One should be able to prove interesting structure theorems for such manifolds. For instance, I expect that the abelian category of analytic coherent sheaves on such manifolds is stable under deformation equivalence, i.e. if two such manifolds are deformation equivalent, they should have equivalent categories of coherent sheaves.</p> <p>The one danger here is that the pool of examples may be very small. I guess, the first thing to look at is to find constructions of more examples. It will be interesting to find an example with a non-abelian infinte fundamental group. One can try to take a quotient of a torus by a freely acting finite group but I suspect that these are never simple.</p> <p>Also, as Dmitry points out, the terminology is misleading, so if you are seriously thinking of working on these manifolds, now is the time to come up with a better name for them.</p> http://mathoverflow.net/questions/26446/references-for-complex-analytic-geometry/26476#26476 Answer by Tony Pantev for References for complex analytic geometry? Tony Pantev 2010-05-30T18:06:35Z 2010-05-31T02:14:54Z <p>Two books that I like a lot:</p> <p>1) Joseph Taylor's <a href="http://books.google.com/books?id=i8lUNpZ379MC&amp;printsec=frontcover&amp;dq=Taylor,+Several+complex+variables&amp;source=bl&amp;ots=7MVn4Xlu9E&amp;sig=Bwt3jHAJvq5LyuuW8TODoykm9eE&amp;hl=en&amp;ei=0KcCTIL6LsaqlAeEwtGiCA&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CBYQ6AEwAA#v=onepage&amp;q&amp;f=false" rel="nofollow"> Several complex variables with connections to algebraic geometry and Lie groups </a>.</p> <p>2) Constantin Banica and Octavian Stanasila's "Algebraic methods in the global theory of complex spaces" , Wiley (1976) </p> http://mathoverflow.net/questions/23614/math-history-books/23698#23698 Answer by Tony Pantev for Math History books Tony Pantev 2010-05-06T12:34:56Z 2010-05-06T12:34:56Z <p>I will add my vote for John Stillwell's book. It has a wealth of material to draw on, and it is gorgeously written.</p> <p>Take also a look at "Geometry for the liberal arts" by Dan Pedoe. It is mainly about geometry but has a lot of Renaissance examples. It also has exercises, which is certainly a plus for your purposes.</p> http://mathoverflow.net/questions/18631/mirror-of-local-calabi-yau/18711#18711 Answer by Tony Pantev for Mirror of local Calabi-Yau Tony Pantev 2010-03-19T02:55:46Z 2010-03-30T20:27:09Z <p>The physicists (see e.g. <a href="http://arxiv.org/abs/hep-th/0012041" rel="nofollow"> this paper of Aganagic and Vafa</a>) will write the mirror as a threefold $X$ which is an affine conic bundle over the holomorphic symplectic surface $\mathbb{C}^{\times}\times \mathbb{C}^{\times}$ with discriminant a Seiberg-Witten curve $\Sigma \subset \mathbb{C}^{\times}\times \mathbb{C}^{\times}$. In terms of the affine coordinates $(u,v)$ on $\mathbb{C}^{\times}\times \mathbb{C}^{\times}$, the curve $\Sigma$ is given by the equation $$\Sigma : \ u + v + a uv^{-1} + 1 = 0,$$ and so $X$ is the hypersurface in $\mathbb{C}^{\times}\times \mathbb{C}^{\times} \times \mathbb{C}^2$ given by the equation $$X : \ xy = u + v + a uv^{-1} + 1.$$</p> <p>From geometric point of view it may be more natural to think of the mirror not as an affine conic fibration over a surface but as an affine fibration by two dimensional quadrics over a curve. The idea will be to start with the Landau-Ginzburg mirror of $\mathbb{P}^{1}$, which is $\mathbb{C}^{\times}$ equipped with the superpotential $w = s + as^{-1}$ and to consider a bundle of affine two dimensional quadrics on $\mathbb{C}^{\times}$ which degenerates along a smooth fiber of the superpotential, e.g. the fiber $w^{-1}(0)$. In this setting the mirror will be a hypersurface in $\mathbb{C}^{\times}\times \mathbb{C}^{3}$ given by the equation $$xy - z^2 = s + as^{-1}.$$ Up to change of variables this is equivalent to the previous picture but it also makes sense in non-toric situations. Presumably one can obtain this way the mirror of a Calabi-Yau which is the total space of a rank two (semistable) vector bundle of canonical determinant on a curve of higher genus. </p> http://mathoverflow.net/questions/18326/positivity-in-stack-geometry/18335#18335 Answer by Tony Pantev for Positivity in stack geometry Tony Pantev 2010-03-16T04:18:01Z 2010-03-16T12:02:40Z <p>Perhaps this <a href="http://www.math.uzh.ch/fileadmin/user/kresch/publikation/geodm.pdf" rel="nofollow"> paper of Andrew Kresch</a> will be helpful. He discusses the notion of a quasi-projective DM stack at length and gives several characterizations of this type of quasi-projectivity.</p> http://mathoverflow.net/questions/17578/triangulating-surfaces/17582#17582 Answer by Tony Pantev for Triangulating surfaces Tony Pantev 2010-03-09T06:13:54Z 2010-03-09T06:32:23Z <p>Try this <a href="http://www.cis.upenn.edu/~jean/gbooks/surftop.html" rel="nofollow">book</a> by Jean Gallier and Diana Xu. It is aimed at undergraduates and has a nice account of Thomassen's elementary proof of the triangulation theorem in the last appendix. Or you can refer the students to Thomassen's original paper which is also quite readable.</p> http://mathoverflow.net/questions/14861/is-there-a-refinement-of-the-hochschild-kostant-rosenberg-theorem-for-cohomology/14911#14911 Answer by Tony Pantev for Is there a refinement of the Hochschild-Kostant-Rosenberg theorem for cohomology? Tony Pantev 2010-02-10T15:47:58Z 2010-02-11T01:41:37Z <p>Yes there is. It was noted by Kontsevich long time ago that the HKR quasi-isomorphism on cochains can be corrected to give a quasi-isomorphism of dg-algebras and thus induce an $A_\infty$ quasi-isomorphism of minimal models. The correction is very natural - one needs to compose the HKR map it the contraction by the square root of the Todd class, where the latter is understood as a polynomial of the Atiyah class. This story has been studied in great detail in the past few years and has been generalized further to give Tsygan formality which is a quasi-isomorphism of $\infty$-calculi. This was proven by Dolgushev-Tamarkin-Tsygan and also by Calaque-Rossi-van den Bergh.</p> <p>The literature on the subject is huge but you should get a good sense of the results if you look at this <a href="http://arxiv.org/abs/0901.0069" rel="nofollow">survey</a> by Dolgushev-Tamarkin-Tsygan and at this <a href="http://arxiv.org/abs/0904.4890" rel="nofollow">paper</a> of Calaque-Rossi-van den Bergh. There are also many interesting references listed in these papers, for instance the works of Caldararu on the Mukai pairing.</p> http://mathoverflow.net/questions/1912/properties-of-monodromy-of-a-fibration/2025#2025 Answer by Tony Pantev for Properties of monodromy of a fibration? Tony Pantev 2009-10-23T03:29:27Z 2010-01-15T08:52:41Z <p>A <strong>small clarification</strong> on bhargav's answer: in algebraic geometry we only have quasi-unipotency of the <em>local</em> monodromy in one-parameter families (which is what bhargav is talking about); or in multi-parameter families but only near a normal crossing point of the discriminant. Global monodromies are reductive and local monodromies near bad points of the discriminant can be more general.</p> <p><strong>For concreteness</strong> look at a projective morphism $f : X \to B$, where $X$, $B$ are smooth complex projective varieties. Let $D \subset B$ be the discriminant divisor of $f$, i.e. the divisor where the differential of $f$ is not surjective. The global monodromy of the smooth fibration $f : X - f^{-1}(D) \to B - D$ is always reductive by a theorem of Borel. That is: the Zariski closure of the monodromy in the linear automorphisms of the cohomology of the marked fiber is a complex reductive group. If we take a small analytic ball $U \subset B$ centered at some point of $D$, and if we know that $D\cap U$ is a normal crossings divisor in $U$, then the monodromy of the local fibration $f : f^{-1}(U-D) \to U-D$ is quasi-unipotent as bhargav explained. Note that the normal crossings condition implies that the fundamental group of $U - D$ is abelian, so the quasi-unipotency condition makes sense here. </p> <p>If however $U\cap D$ does not have normal crossings, then $\pi_{1}(U-D)$ need not be abelian and the monodromy of $f : f^{-1}(U-D) \to U-D$ need not be quasi-unipotent. An easy example is to look at a generic projective plane $\mathbb{P}^{2}$ in the $9$ dimensional projective space of cubic curves in $\mathbb{P}^{2}$. This plane parametrizes a family of cubics which degenerates along a discriminant curve $D \subset \mathbb{P}^{2}$ and under the genericity assumption $D$ has only nodes and cusps. The cuspidal points of $D$ correspond to cuspidal cubics, and near a cusp of $D$ the local fundamental group of $U - D$ is the amalgamated product of $\mathbb{Z}/4$ and $\mathbb{Z}/6$ over $\mathbb{Z}/2$ and so is isomorphic to $SL_{2}(\mathbb{Z})$ the local monodromy representation near the cusp, i.e. the representation of $\pi_{1}(U-D,u_{0})$ to the linear automorphisms of the first integral cohomology of the cubic corresponding to $u_{0} \in U - D$, is an inclusion, i.e. has image $SL_{2}(\mathbb{Z})$. In particular it is not quasi-unipotent.</p> http://mathoverflow.net/questions/11774/difference-between-equivalence-relations-on-algebraic-cycles/11805#11805 Answer by Tony Pantev for difference between equivalence relations on algebraic cycles Tony Pantev 2010-01-15T00:33:00Z 2010-01-15T00:33:00Z <p>It is indeed true that rational equivalence gives bigger groups of cycles than say algebraic equivalence. However algebraic equivalence is also far away from homological equivalence. In complex geometry people often study a basic invariant of a variety $X$ called the <em>Griffiths group</em>. By definition the Griffiths group $Gr(X)$ is the group of cycles homologous to zero (in the classical topology) modulo cycles algebraically equivalent to zero. Griffiths originally showed that this group can contain non-torsion elements, and Clemens showed that it can happen that $Gr(X)\otimes \mathbb{Q}$ is infinite dimensional as a rational vector space. People have studied Griffiths groups quite a bit and have proven some great theorems about them. For instance Voisin showed that the Griffiths group of a Calabi-Yau threefold which is general in its moduli is infinitely generated. </p> http://mathoverflow.net/questions/10631/homology-class-orthogonal-to-image-of-chern-characters/10660#10660 Answer by Tony Pantev for Homology class orthogonal to image of Chern characters? Tony Pantev 2010-01-04T04:00:12Z 2010-01-04T04:00:12Z <p>Why not just take a class that is orthogonal to all the algebraic classes on $X$? For instance you can take $Y$ to be a point, and take $X$ to be a generic abelian surface over $\mathbb{C}$, i.e. and abelian surface with $NS(X) = \mathbb{Z}$. The Chern character of any coherent sheaf on $X$ is contained in $H^{0}(X)\oplus NS(X)\oplus H^{4}(X)$. Take now any $t \in NS(X)^{\perp} \subset H^{2}(X)$, i.e. a transcendental cohomology class of degree two.</p> http://mathoverflow.net/questions/10212/moduli-spaces-of-coherent-sheaves-on-k3s/10220#10220 Answer by Tony Pantev for Moduli spaces of coherent sheaves on K3s Tony Pantev 2009-12-31T03:03:18Z 2009-12-31T03:03:18Z <p>This follows from a result of Yoshioka. In Theorem 8.1 of this <a href="http://arxiv.org/abs/math/0009001" rel="nofollow">paper</a> Yoshioka showed that every moduli space of coherent sheaves on a K3 surface $X$ is deformation equivalent to an appropriate Hilbert scheme of points of $X$. Since every K3 is deformation equivalent to an elliptic K3 it follows that their Hilbert schemes are deformation equivalent and so you get the statement that you wanted.</p> http://mathoverflow.net/questions/107841/for-a-hyperplane-section-z-of-x-when-there-exists-its-etale-x-neighbourhood-such Comment by Tony Pantev Tony Pantev 2012-09-22T14:51:55Z 2012-09-22T14:51:55Z I am not sure I understand the question. If you had such an etale neighborhood of $Z$, then the normal bundle of $Z$ in $U$ will be trivial. But the normal bundle to $Z$ in $U$ is isomorphic to the normal bundle of $Z$ in $X$, and $N_{Z/X}$ can never be trivial for a hyperplane section. http://mathoverflow.net/questions/106206/singular-fibers-of-an-elliptic-fibered-k3-surface Comment by Tony Pantev Tony Pantev 2012-09-03T01:50:22Z 2012-09-03T01:50:22Z If $S$ is elliptic, then the fiber and the section generate a sublattice in $Pic(S)$ which is isomorphic to $U$. So if $S$ is elliptic and $Pic(S) \cong U(k)$ you will have $k = 1$. Did you mean to ask for $S$ to be only genus one fibered and $Pic(S) \cong U(k)$ ? http://mathoverflow.net/questions/102802/quasi-unipotent-monodromy-for-general-families/102847#102847 Comment by Tony Pantev Tony Pantev 2012-07-23T03:07:50Z 2012-07-23T03:07:50Z Two classical references are the paper &quot;Periods of integrals on algebraic manifolds III&quot;, Publ. Math. IHES 38 (1970) by Griffiths and the paper &quot;Variation of Hodge structure: the singularities of the period mapping&quot; Invent. math. 22 (1973) by Schmid. They in particular explain Borel's proof of the quasi-unipotency theorem that I mentioned above. There are many other modern references. For instance, you may want to take a look at the excellent book &quot;Period mappings and period domains&quot; by Carlson, Mueller-Stach, and Peters. http://mathoverflow.net/questions/102802/quasi-unipotent-monodromy-for-general-families/102847#102847 Comment by Tony Pantev Tony Pantev 2012-07-23T03:00:56Z 2012-07-23T03:00:56Z Ah, I noticed I missed an adjective in the comment - I was defining what it means for $G$ to be quasi-unipotent. I edited the answer to reflect this correctly. The unipotency of $mon$ is defined in a similar manner - we say that $mon$ is unpotent, when $G$ is a connected unipotent algebraic group, i.e. when when it coincides with its unipotent radical. The references are numerous and the applications are usually Hodge theoretic. The quasi-unipotency of a local systen can is also very useful when we compute cohomology. http://mathoverflow.net/questions/99200/question-on-k3-surface/99216#99216 Comment by Tony Pantev Tony Pantev 2012-06-10T18:50:16Z 2012-06-10T18:50:16Z @alex24: You should accept Remke's answer. It correctly and completely answers your question. http://mathoverflow.net/questions/99200/question-on-k3-surface/99216#99216 Comment by Tony Pantev Tony Pantev 2012-06-10T18:43:40Z 2012-06-10T18:43:40Z As Remke Kloosterman points out - I got my Hurwitz formula wrong. To get the canonical class of the double cover to be trivial, you need to take a branch divisor that is in the twice the anticanonical linear system of the rational elliptic surface. So you simply need to take two smooth fibers as your branch divisor. This gives a smooth K3. It is the fiber product of the rational elliptic surface and a $\mathbb{P}^{1}$ doubly covering your original $\mathbb{P}^{1}$ with branching at the two points over which the two smooth fibers sit. http://mathoverflow.net/questions/99200/question-on-k3-surface/99226#99226 Comment by Tony Pantev Tony Pantev 2012-06-10T18:36:18Z 2012-06-10T18:36:18Z Of course, you are right! I don't know what I was thinking. Somehow I did the calculation in my head without writing and, of course, I got it wrong. http://mathoverflow.net/questions/23614/math-history-books/23698#23698 Comment by Tony Pantev Tony Pantev 2011-09-10T03:16:22Z 2011-09-10T03:16:22Z Actually Pedoe's book has several editions. The first edition is called &quot;Geometry for the liberal arts&quot; and the second is called &quot;Geometry and the visual arts&quot;. Except for the title the two editions seem to be identical. Our library has a copy only of the first edition but the second edition is still in print so it may be easier to find. http://mathoverflow.net/questions/73483/exact-sequences-of-bundles-on-grassmannians Comment by Tony Pantev Tony Pantev 2011-08-23T12:17:14Z 2011-08-23T12:17:14Z I may be missing something but it seems that you may be able to get these sequences by applying the standard Schur complex functors (as in section 2.4 of J.Weyman's book &quot;Cohomology of vector bundles and syzygies) to the tautological short exact sequence on the Grassmanian. http://mathoverflow.net/questions/72998/finite-fundamental-groups-of-3-dimensional-calabi-yau-manifolds/73005#73005 Comment by Tony Pantev Tony Pantev 2011-08-17T01:52:56Z 2011-08-17T01:52:56Z There is a very nice and detailed description of this CY in the Gross-Pavanelli paper <a href="http://arxiv.org/abs/math/0512182" rel="nofollow">arxiv.org/abs/math/0512182</a>. If you have not seen those, you may also want to take a look at this paper <a href="http://arxiv.org/abs/math/0609728" rel="nofollow">arxiv.org/abs/math/0609728</a> by Borisov-Hua, and the paper <a href="http://arxiv.org/abs/math/0609728" rel="nofollow">arxiv.org/abs/math/0609728</a> of Bouchard-Donagi. http://mathoverflow.net/questions/72998/finite-fundamental-groups-of-3-dimensional-calabi-yau-manifolds/73005#73005 Comment by Tony Pantev Tony Pantev 2011-08-17T00:47:45Z 2011-08-17T00:47:45Z There are many possible constructions of the first one: via toric geometry, via elliptic fibrations, via abelian surface fibrations, etc. The elliptic fibration construction is written for instance in my old paper &lt;a href=&quot;<a href="http://arxiv.org/abs/hep-th/0410055&quot;&gt;http://arxiv.org/abs/hep-th/0410055&lt;/a&gt" rel="nofollow">arxiv.org/abs/hep-th/0410055&quot;&gt;http://&hellip;</a>;. You can see there that the group acts freely on the total space but acts with fixed points on the base of the elliptic fibration. The second Calabi-Yau was originally constructed by Gross-Popescu as a pencil of abelian surfaces with polarizations $(1,8)$. http://mathoverflow.net/questions/68356/orthogonal-complements-of-root-lattices-in-e-8/68368#68368 Comment by Tony Pantev Tony Pantev 2011-06-21T14:06:42Z 2011-06-21T14:06:42Z The first time I encountered all this was in the excellent paper: Oguiso, Keiji; Shioda, Tetsuji The Mordell-Weil lattice of a rational elliptic surface. Comment. Math. Univ. St. Paul. 40 (1991), no. 1, 83–99. Among other things they show that both embeddings of all five exceptions do occur as narrow Mordell-Weil lattices of rational elliptic surfaces. http://mathoverflow.net/questions/68356/orthogonal-complements-of-root-lattices-in-e-8/68368#68368 Comment by Tony Pantev Tony Pantev 2011-06-21T14:06:33Z 2011-06-21T14:06:33Z The original paper is: Dynkin, E. B. Semisimple subalgebras of semisimple Lie algebras. (Russian) Mat. Sbornik N.S. 30(72), (1952), 349–462. There is also an English translation: E.B. Dynkin, Semisimple subalgebras of semisimple Lie algebras. AMS Translations, volume 6, (1957), 111-244. The part you need is in Table 11 in Chapter II. http://mathoverflow.net/questions/68356/orthogonal-complements-of-root-lattices-in-e-8/68368#68368 Comment by Tony Pantev Tony Pantev 2011-06-21T13:09:46Z 2011-06-21T13:09:46Z $A_{1}^{\oplus 4}$, ($D_{4}$, $A_{1}^{\oplus 4}$); \linebreak $A_{3}\oplus A_{1}^{\oplus 2}$, ($A_{3}$, $A_{1}^{\oplus 2}\oplus \langle 4\rangle$); \linebreak $A_{3}^{\oplus 2}$, ($A_{1}^{\oplus 2}$, $\langle 4 \rangle^{\oplus 2}$), \linebreak $A_{5}\oplus A_{1}$, ($A_{2}$, $A_{1}\oplus \langle 6 \rangle$); \linebreak $A_{7}$, ($A_{1}$, $\langle 8\rangle$). \linebreak So for the primitive embeddings you get a uniquely determined orthogonal complement. http://mathoverflow.net/questions/68356/orthogonal-complements-of-root-lattices-in-e-8/68368#68368 Comment by Tony Pantev Tony Pantev 2011-06-21T13:00:22Z 2011-06-21T13:00:22Z According to Dynkin's classification of root sublattices in $E_{8}$, there are only five root subblattices in $E_{8}$ (modulo the Weyl group action) that admit more than one embedding in $E_{8}$. Moreover each of the five admits exactly two non-equivalent embeddings. The five exceptions (and their possible complements) are: